Let $S$ be a commutative ring, let $G$ be a finite group acting on $S$ via automorphisms (not necessarily faithfully), and let $R$ be a subring of $S$ consisting of elements fixed $G$. The extension $S/R$ is said to be $G$-Galois if there exists a faithfully flat commutative $R$-algebra $R'$ such that $S\otimes_RR'\cong \prod_{g\in G}R'$ as $R'$-algebras carrying a $G$-action. Weaking this requirement significantly, call $S/R$ a quasi-$G$-Galois extension if $R$ consists of all elements of $S$ fixed by $G$ and $S$ is a projective $R$-module of rank $|G|$.
When $S/R$ is quasi-$G$-Galois, we can define two norms from $S$ to $R$. The first is the ordinary norm, $N_{S/R}$, sending $a\in S$ to the determinant of $[x\mapsto ax]\in \mathrm{End}_R(S)$. The second is the $G$-norm, defined by $N_G(s)=\prod_{g\in G}g(s)$.
It is well-known that $N_G=N_{S/R}$ when $S/R$ is $G$-Galois (it is enough to check this after a faithfully flat base change, so one reduces to the case $S=\prod_{g\in G}R$). My question is whether $N_G=N_{S/R}$ when $S/R$ is quasi-$G$-Galois.
This is nontrivial even when $R$ is a field. For instance, let $G=\langle g|g^4=1\rangle$ act on $S:=\mathbb{C}[x,y,z]/(x,y,z)^2$ by letting $g$ fix $\mathbb{C}$ and send $x$, $y$ and $z$ to their negatives ($G$ does not act faithfully). One readily checks that $R:=\mathbb{C}$ is the subring of elements fixed $G$, and $S$ is free of rank $|G|=4$ over it. Direct computation shows that indeed $N_G=N_{S/R}$.
Sufficient conditions on $R$, $S$ and $G$ that guarantee $N_G=N_{S/R}$ are welcome. However, since I already have several of these at the moment, what I am really after is a counterexample or a proof of the general statement.
